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  • There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially, both have equal number of nuclei. After n half lives of A, rate of disintegration of both

There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially, both have
equal number of nuclei. After n half lives of A, rate of disintegration of both are equal. The value of n is:

Option: 1

4


Option: 2

2


Option: 3

1


Option: 4

5


Answers (1)

best_answer

\text { Let } \lambda_{\mathrm{A}}=\lambda

\therefore \quad \lambda_B=2 \lambda

If \mathrm{N}_0 is total number of atoms in \mathrm{A} and \mathrm{B} at \mathrm{t}=0, then initial rate of disintegration of \mathrm{A}=\lambda \mathrm{N}_0, and initial rate of disintegration of \mathrm{B}=2 \lambda \mathrm{N}_0 .

\text { As } \lambda_B=2 \lambda_A                       \left(\because \lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}\right)

\therefore \quad\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{B}}=\frac{1}{2}\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{A}}

i.e., half – life of B is half the half – life of A

After one half – life of A

       \left(-\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{\mathrm{A}}=\frac{\lambda \mathrm{N}_0}{2}                                (i)

Equivalently, after two half lives of B

       \left(-\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{\mathrm{B}}=\frac{2 \lambda \mathrm{N}_0}{4}=\frac{\lambda \mathrm{N}_0}{2}                (ii)

From (i) and (ii), we get

\left(-\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{\mathrm{A}}=\left(-\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{\mathrm{B}}

After n = 1, i.e., one half – life of A, the rate of disintegration of both will be equal.

Posted by

Ritika Jonwal

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